Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define fill( x, y ) memset( x, y, sizeof x )
#define copy( x, y ) memcpy( x, y, sizeof x )
using namespace std;
 
typedef long long LL;
typedef pair < int, int > pa;
 
inline int read()
{
	int sc = 0, f = 1; char ch = getchar();
	while( ch < '0' || ch > '9' ) { if( ch == '-' ) f = -1; ch = getchar(); }
	while( ch >= '0' && ch <= '9' ) sc = sc * 10 + ch - '0', ch = getchar();
	return sc * f;
}
 
const int MAXN = 200020;
 
int q[MAXN], p[MAXN], n, T[MAXN], f[MAXN];
 
int main()
{
#ifdef wxh010910
	freopen( "data.in", "r", stdin );
#endif
	n = read();
	for( int i = 1 ; i <= n ; i++ ) q[ p[ i ] = read() ] = i;
	for( int i = n - 1 ; i ; i-- )
	{
		if( !T[ i + 1 ] )
		{
			if( q[ i ] > q[ i + 1 ] ) T[ i ] = 1, f[ i ] = i + 1;
			else T[ i ] = 0;
		}
		else
		{
			int cnt = 0;
			cnt += q[ f[ i + 1 ] ] < q[ i ];
			cnt += q[ i ] < q[ i + 1 ];
			cnt += q[ i + 1 ] < q[ f[ i + 1 ] ];
			if( cnt == 2 ) T[ i ] = T[ i + 1 ], f[ i ] = f[ i + 1 ];
			else T[ i ] = T[ i + 1 ] + 1, f[ i ] = i + 1;
		}
	}
	return printf( "%d\n", T[ 1 ] ), 0;
}